physics
posted by Ray Dunnigan .
A stone is dropped from the roof of a high building. A second stone is dropped 1.74 seconds later.
Pt. A) How far apart are the stones when one has reached a speed of 13.2 m/s^2?

The second stone reached the velocity 13.2 m/s for the time
t2=v/g=13.2/9.8 =1.35 s.
The distance covered by the second ball for this time is
h2= =g•t2²/2 =9.8•(1.35) ²/2 =8.9 m
The first stone moved during
t=t1+t2 =1.74+1.35 =3.09 s.
The distance covered by the first stone for 3.09 s is
h1=g•t1²/2 =9.8•(3.09) ²/2 =46.8 m.
Δh=h1h2= 46.88.9= 37.9 m.
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